`int_(0)^(1) lnsin(pi/2x) dx`

Evaluate int_(0)^(pi//2)lnsin2xdx

Evaluate int_(0)^(pi//2)lnsin2xdx

If int_(0)^(pi//2) ln (sin x) dx= - pi/2 ln 2 then int_(0)^(pi) ln (1+ cos x) dx=

int_(0)^(1) sin^(-1) x dx =(pi)/(2) -1

int_(0)^(1) sin^(-1) x dx =(pi)/(2) -1

Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2

show that (a) int_(0) ^(2pi) sin ^(3) x dx = 0 , (b) int_(-1)^(1) e^(-x^(2)) dx = 2 int_(0)^(1) e^(-x^(2)) dx

I_(1)=int_(0)^((pi)/2)(sinx-cosx)/(1+sinxcosx)dx, I_(2)=int_(0)^(2pi)cos^(6)dx , I_(3)=int_(-(pi)/2)^((pi)/2)sin^(3)xdx, I_(4)=int_(0)^(1) In (1/x-1)dx . Then